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Dmitriy789 [7]
2 years ago
8

4p + 5c = p. Solve for c

Mathematics
2 answers:
PilotLPTM [1.2K]2 years ago
4 0

Answer:

4p + 5c = p

4p - 4p + 5c = p-4p

5c = -3p

5c/5 = -3p/5

c = -3p/5

hope that will help you

madam [21]2 years ago
3 0
Minus 4p on both sides
you end up with 5c = -3p
divide by 5 on both sides
c = -3/5p
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Yuliya22 [10]

Your answer is the last option, 6.


We can work this out by looking for the total of people that like tennis in the table, and subtracting it from the total amount of people that like baseball.


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I hope this helps!

3 0
3 years ago
Someone please help me on this!
Alexxx [7]

Answer & Step-by-step explanation:

This can be proven with the SAS theorem (side-angle-side)

With a perpendicular bisector, the line it bisects is cut directly in half. This creates two equal sides:

GJ=IJ

and it creates two 90° angles:

∠GJH=∠IJH

And because of the reflexive property of congruence:

JH=JH

Side-Angle-Side.

:Done

6 0
3 years ago
Quadrilateral ABCD is inscribed in a circle. What is the measure of angle A? Enter your answer in the box.
SashulF [63]

Answer:

135°

Step-by-step Explanation:

==>Given:

An inscribed quadrilateral ABCD with,

m<A = (3x +6)°

m<C = (x + 2)°

==>Required:

measure of angle A

==>Solution:

First, let's find the value of x.

Recall that the opposite angles in any inscribed quadrilateral in a circle are supplementary.

Therefore, this means m<A + m<C = 180°

Thus, (3x+6) + (x+2} = 180

3x + 6 + x + 2 = 180

Collect like terms:

3x + x + 6 + 2 = 180

4x + 8 = 180

Subtract 8 from both sides:

4x + 8 - 8 = 180 - 8

4x = 172

Divide both sides by 4:

4x/4 = 172/4

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We can now find m<A = (3x + 6)°

m<A = 3(43) + 6

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6 0
3 years ago
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Molodets [167]

Answer:

\lim_{x \to 2^-} \frac{|x-2|}{x-2}=-1.

Step-by-step explanation:

We want to find \lim_{x \to 2^-} \frac{|x-2|}{x-2}.

By definition:

|x-2|=\left \{ {{x-2,\:if\:x\:>\:2} \atop {-(x-2),\:if\:x\:

Since we want to find the Left Hand Limit, we use f(x)=-(x-2)

\implies \lim_{x \to 2^-} \frac{|x-2|}{x-2}=\lim_{x \to 2} \frac{-(x-2)}{x-2}.

\implies \lim_{x \to 2^-} \frac{|x-2|}{x-2}=\lim_{x \to 2} (-1).

The limit of a constant is the constant.

\implies \lim_{x \to 2^-} \frac{|x-2|}{x-2}=-1.

8 0
3 years ago
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aleksandrvk [35]

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